Answer:
PROGRAM QuadraticEquation
Solver
IMPLICIT NONE
REAL :: a, b, c
;
REA :: d
;
REAL :: root1, root2
;
//read in the coefficients a, b and c
READ(*,*) a, b, c
WRITE(*,*) 'a = ', a
WRITE(*,*) 'b = ', b
WRITE(*,*) 'c = ', c
WRITE(*,*)
// computing the square root of discriminant d
d = b*b - 4.0*a*c
IF (d >= 0.0) THEN //checking if it is solvable?
d = SQRT(d)
root1 = (-b + d)/(2.0*a) // first root
root2 = (-b - d)/(2.0*a) // second root
WRITE(*,*) 'Roots are ', root1, ' and ', root2
ELSE //complex roots
WRITE(*,*) 'There is no real roots!'
WRITE(*,*) 'Discriminant = ', d
END IF
END PROGRAM QuadraticEquationSolver
Answer and Explanation:
In C programming language:
char fun(int*a, int*b){
printf("enter two integers: ");
scanf("%d%d",a,b);
int e;
printf("please enter a character: ");
e=getchar();
return e;
}
int main(int argc, char *argv[]) {
int d;
int f;
int g;
fun();
printf("%d%d%d", d, f, g);
}
We have declared a function fun type char above and called it in main. Note how he use the getchar function in c which reads the next available character(after the user inputs with printf()) and returns it as an integer. We then return the variable e holding the integer value as char fun() return value.
Answer:
The correct answer to the following question is options A, B, and D.
Explanation:
Base Transceiver Station (BTS) is an apparatus that works wireless communication between the network and user equipment (UE - Mobile phones (handset), wireless network computers). Wireless technologies like wi-fi, GSM, CDMA, etc.
Some elements of BTS :
- Radio base station
- Tower/Mast
- Duplexer
- Transceiver unit (TRU)
- Microwave
So the above are the elements of the BSA, then it also having the characteristics related to these elements like it definitely connects to a cell network, sends phone or mobile phone and microwave signals to cellular providers but it isn't able to send the signals to the recepients.
Answer: True
Explanation:
A language is said to be closed under a operation here the complement is the operation then if upon application of that operation to any members of that language always yields a member of that language.
regular languages are closed under complement. A proof of the statement is
If a regular language 'L' is regular then there is a DFA X recognizing that regular language 'L'. to show that L' (compliment) is regular we need to have another DFA X' recognizing L'.
The initial state and transition function of both the DFAs are same except their accepting state. Then we can say that X' accepts L'.
So, we can say that regular languages are closed under complement.
A. fuel ; when you but a car you aren't buying the gas to go in it
